$h(x) = \begin{cases} x^3&, & x\in (-\infty,0] \\\\ \dfrac{24}{x}-1 &, & x \in (0,8]\\\\ (x-1)(x+3) &, & x \in (8,\infty)\end{cases}$ $h(-3)=$
The strategy First, we should find the appropriate assignment rule out of the three, by checking which case applies for $x={-3}$. [I don't understand the notation for the cases.] Finding the appropriate assignment rule Since ${-3}\in(-\infty,0]$, we should use the first assignment rule $x^3$. The answer $h({-3})=({-3})^3=-27$ In conclusion, $h(-3)=-27$.